Why doesn’t a transformer break the conservation of energy? The way I see it, using a step-up transformer, you can up the volts to whatever you want…and then the amps can be whatever you want in that higher voltage line based on its resistance. So, is this free energy? Of course not, but why not?
I’m not an EE, so go easy on me…! 
- Jinx
No. In an ideal transformer, the voltage across the primary equals the voltage across the secondary times the turns ratio of the primary to the secondary. However, the current through the primary equals the current through the secondary divided by the turns ratio. In other words, the energy in equals the energy out, exactly. In a real transformer, there are also losses in the copper and iron that mean the energy out is slightly less than the energy in.
If you transmit 100 amperes at 110,000 volts, you’ve sent 11 megawatts down the line. If that arrives at a transformer, and gets turned into 100,000 amperes at 110 volts, it’s the same 11 megawatts. The wattage, or amount of actual energy transmitted, remains the same.
Wattage is a measurement of power, not energy. But, your point is valid, because energy equals power multiplied by time, and transformers don’t alter time. Yet. 
Was I the only one who thought this was going to be about these Transformers and the conservation of energy? 
Hey, when I saw the thread title I was going to ask why he’s worrying about this when elderly transformers have beards and they cough when exposed to polluted air!
There’s additional losses due to (1) hysteresis which is the metal’s incomplete changing from north poles to south poles and back 120 times a second and (2) eddy currents which are induced currents opposing the flow of the desired current.
There’s little tricks to reduce both of these, too.
Nope. I opened this thread just to see how many posts it would take for someone to make a Starscream joke and I’m very disappointed.
Those are some of the iron loss components I mentioned. Your definiton of hysteresis is a bit off, though. Hysteresis is really the lagging of the induced B field behind the external magnetizing force, H, and can be plotted on a graph as the familiar (at least to magnetics engineers) B-H curve. Hysteresis can never be completely eliminated. The best one can accomplish is to make the B-H curve as “square” as possible and this is done by tweaking the material composition and grain structure. Eddy currents are reduced by electrically isolating sections of the core from oen another. The smaller the isolated sections, the smaller the eddy currents and therefore, the losses. In standard laminated and tape-wound cores, this is done by making the material as thin as possible, and oxide-coating it to provide electrical insulation. Ferrite cores are ceramics, and as such have low conductivity and correspondingly low eddy-current losses. Reducing hysteresis and eddy-current losses becomes more important with increasing AC frequency. Power-line transformers can make do with fairly thick silicon steel laminations and not suffer great losses; audio transformers must use much thinner laminations or tape-wound cores, using a more square material like one of several nickel alloys; and RF transformers almost always use one of the soft ferrites.
Would it be pathetic and tiresome of me to insert a Transformers joke at this point?
You really are oversimplifying the mechanics of the conversion. There really is more to it than meets the eye.
I didn’t think so either.
That should more properly read “…the induced magnetic flux, B…”
Actually the conservation of energy is used to work out the voltage and current relationships in primary and secondary of a transformer.
The primary and secondary voltages are related by the turns ratio:
E[sub]p[/sub]/E[sub]s[/sub] = N[sub]p[/sub]/N[sub]s[/sub]
The transformer is to a good approxumation a lossless device and by the conservation of energy the primary power must equal the secondary power.
For a resistive load this is: E[sub]s[/sub]*I[sub]s[/sub] = E[sub]p[/sub]*I[sub]p[/sub]
or: I[sub]s[/sub]/I[sub]p[/sub] = E[sub]p[/sub]/E[sub]s[/sub]
Now say its a step down transformer with the turns ratio of N[sub]p[/sub]/N[sub]s[/sub] = 10
Then: E[sub]p[/sub]/E[sub]s[/sub] = 10 and I[sub]s[/sub]/I[sub]p[/sub] =10
So the voltage is been stepped up by 10 and the current has been stepped down by 10.
For an actual transformer these relationships for an ideal transformer are inserted into an equivalent circuit that involves resistive losses to account for winding resistance and core losses. If the frequency response of the transformer is important, such as for audio transformers, the inductance and capacitance of the device is also included in the equivalent circuit.
Let’s make this sentence “So the voltage is been stepped up by 10 and the current has been stepped down by 10.” read: So the voltage is been stepped down by 10 and the current has been stepped up by 10.
I dunno… I kind of stand by my simplified hysteresis loss explaination. And I’m not sure what you mean by a ‘square’ B-H curve being desireable, either square in the non-linear, mathmatical sense or the graphical sense with respect to the shape of the curve. The ideal B-H graph would be a thin line approximating the overall shape of the hysteresis curve.
Well, ya, but lagging implies time and time isn’t part of it. The little molecular magnetic dipoles don’t align back and forth without losses.
I agree, for use in transformers the ideal magnetization curve would be one with no hysteresis or non-linearity. A square hysteresis loop would be useful for things like magnetic memory cores, etc.
I didn’t say a square B-H curve is ideal, I said it’s the best you can do with real-world cores. You’re right that ideally, the curve would be linear, but that’s never going to happen.
Incidently, or not so incidently, the assumption of the transformer as a lossless device, thus allowing use of the conservation of energy to equate power in to power out, isn’t just pulled out of the air.
An ideal transformer is merely two inductances coupled together. Inductances are shown by theory and experiment to be lossless elements, storing energy in a magnetic field. Losses in actual physical inductances are the result of unavoidable resistance in the windings and not a result of the inductance.
You know, I almost said non-linear in describing the ideal curve but it depends on whether your ideal model can saturate or not. I went with ideal coercivity but kept the saturation.
Different ideal transformer models for different folks, I alway say.
This isn’t a particularly good explanation. Hysteresis can be eliminated by using a non-ferromagnetic core, and the “squaring” of the B-H curve properly refers only to something one would do to simplify the mathematical model of a real transformer. A square B-H curve has infinite permeability up to a point, implying saturation, and also a non-zero width, implying hysteresis loss. Certain ferromagnetic materials have properties approaching “square”, but they’re by no means ideal materials for a transformer.
In truth, the best one can do to eliminate hysterisis by tweaking materials is not to “square up” the curve, but to narrow it. Using a non-ferromagnetic material will narrow the hysteresis curve to zero width, but, this might not be ideal for other reasons.
Ferromagnetic materials can be broadly separated in to “soft” and “hard” categories. “Soft” ferromagnetic materials, such as pure iron or iron silicon alloys, don’t make good permanent magnets. As jnglmassiv puts it, they have a low coercivity, which means that it doesn’t take much energy to alter their magnetic orientation. Such materials also necessarily have low magnetic remnance, which is why they don’t make good permanent magnets. But a critical thing here, is that they also have comparitively low magnetic permeabilities.
“Hard” magnetic materials, e.g. steel (iron carbon alloys) or more exotic things like aluminium-nickel-cobalt alloys are the opposite. They make good permanent magnets, and have high coercivity and permeability.
See, there’s a trade-off here. Permeability is good, because it means fewer ampere-turns to create the same magnetic flux, and so less copper is required in the transformer, and so the initial cost of the transformer is lower. Coercivity is bad, because it implies higher hysteresis losses, and could increase both the initial cost of the transformer if extra cooling is required, and the running cost of the transformer over its lifetime, because of the cost of the electricity it’s wasting.
Regards,
Des
Who doesn’t actually design transformers, and who has no interest in doing so in the future.
David Simmons is talking about an “ideal” transformer. The model is perfectly straightforward. The B-H curve is a vertical line coincident with the B axis.
It doesn’t make sense to talk about different “ideal” transformer models. Your model, which includes saturation, isn’t a model of an ideal transformer. It’s an approximation of a real transformer. There’s nothing wrong with that, but you have to recognise that you and David Simmons aren’t talking about the same thing. Q.E.D.'s model included both saturation and hysteresis, and so is another, different, model of a real transformer. Q.E.D. implicitly assumed infinite permeability (i.e., infinite slope on the vertical portions of the curve). You haven’t said what you’ve assumed about permeability in your model.